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Chapter 14: Problem 97

Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{MC}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\).

Short Answer

Expert verified
The pH of the 0.20 M \(C_2H_5NH_2\) solution is 12.23.

Step by step solution

01

Write the base dissociation equation

Ethylamine (C₂H₅NH₂) reacts with water to form the ethylammonium ion (C₂H₅NH₃⁺) and a hydroxide ion (OH⁻). The chemical equation for this reaction is: C₂H₅NH₂ + H₂O ⇌ C₂H₅NH₃⁺ + OH⁻
02

Set up an ICE table

In order to solve for the equilibrium concentration of OH⁻, we'll need to set up an Initial, Change, and Equilibrium (ICE) table for the reaction: ``` [I] 0.20 M - 0 0 [C] -x - +x +x [E] 0.20-x - x x ``` where [I] is the initial concentration, [C] is the amount that changes, and [E] is the equilibrium concentration.
03

Write the Kb expression

In order to find the equilibrium constant for this reaction, we'll need to write an expression for Kb: Kb = \(\frac{[C₂H₅NH₃⁺][OH⁻]}{[C₂H₅NH₂]}\)
04

Substitute the equilibrium concentrations in the Kb expression

Now, we'll substitute the equilibrium concentrations from our ICE table into the Kb expression: \(5.6 × 10^{-4} = \frac{x^2}{0.20-x}\)
05

Solve for the hydroxide ion concentration (x)

Here, we'll assume that x is much smaller than 0.20, meaning that the change in the C₂H₅NH₂ concentration is negligible. This will simplify our reasoning slightly: \(5.6 × 10^{-4} = \frac{x^2}{0.20}\) Now, solve for x: \(x = \sqrt{(5.6 × 10^{-4})(0.20)}\) \(x = 0.0167\) Therefore, [OH⁻] = 0.0167 M.
06

Calculate the pH

To find the pH of the solution, we first need to calculate the pOH using the formula: pOH = -\(\log (0.0167)\) Now, calculate pOH: pOH = 1.77 Finally, we'll use the relationship between pH and pOH (pH + pOH = 14) to find the pH of the solution: pH = 14 - pOH pH = 14 - 1.77 pH = 12.23 The pH of the 0.20 M C₂H₅NH₂ solution is 12.23.

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Most popular questions from this chapter

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\), determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a \(0.100 M\) solution.

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is secreted by ants. Calculate \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of a \(0.025 M\) solution of formic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right)\).

What mass of \(\mathrm{KOH}\) is necessary to prepare \(800.0 \mathrm{~mL}\) of a solution having a \(\mathrm{pH}=11.56\) ?

A typical vitamin C tablet (containing pure ascorbic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\) ) weighs 500 . mg. One vitamin C tablet is dissolved in enough water to make \(200.0 \mathrm{~mL}\) of solution. Calculate the \(\mathrm{pH}\) of this solution. Ascorbic acid is a diprotic acid.
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